Question

Driving Distances. Refer to Exercise 10.48 and determine a 95% confidence interval for the difference between last year's mean VMTs by midwestern and southern households.

Short answer

We can be $95\%$certain that the average VMT of Midwestern households is between $4.69$ and $1.77$ miles less than that of Southern households.

Step-by-step solution

Given Information

Given in the question that, the vehicle miles travelled (VMT) data for two populations of Midwestern and Southern families are provided as samples in the inquiry.

$$\begin{gathered} {\overline{x}}_1=16.23, \\ {s}_1=4.06 \end{gathered}$$

$$\begin{gathered} {\overline{x}}_2=17.69, \\ {s}_2=4.42 \end{gathered}$$

The significance level is 5%.

Explanation

Let's consider the Population 1: Midwestern households:

$$\begin{gathered} {\overline{x}}_1=16.23, \\ {s}_1=4.06 \\ {n}_1=15 \end{gathered}$$

Let's consider the Population 2: Southern households,

$$\begin{gathered} {\overline{x}}_2=17.69, \\ {s}_2=4.42 \\ {n}_2=14 \end{gathered}$$

The main goal is to calculate a $95\%$confidence interval for the difference between two population means,${\mu }_1$and${\mu }_2$

The null and alternate hypotheses should be stated.

Null hypotheses: $H_0:{\mu }_1={\mu }_2$

Alternate hypotheses: $H_a:{\mu }_1\ne {\mu }_2$

Two-tailed hypotheses exist.

Use Table IV to find $t_{\alpha /2}$with $df=n_1+n_2-2$for a confidence level of $1-a$

alpha=0.05 for 95% confidence level,

localid="1651403167353" $$\begin{gathered} \text{df}=n_1+n_2-2 \\ =(15+14-2) \\ =27. \end{gathered}$$

When$\mathrm{df}=27$,using table IV for critical values,

Critical value,

localid="1651403201603" $$\begin{gathered} t_{\alpha /2}=t_{0.05/2} \\ =t_{0.025} \\ =2.052 \end{gathered}$$

Find the confidence interval's endpoints.

Pooled standard deviation:

localid="1651403231074" $s_p=\sqrt{\frac{\left(n_1-1\right)s_1{}^2+\left(n_2-1\right){s_2}^2}{n_1+n_2-2}}$

$s_p=\sqrt{\frac{(15-1)(4.06{)}^2+(14-1\left)\right(4.42{)}^2}{15+14-2}}$

$s_p=\sqrt{\frac{14(16.4836)+13(19.5364)}{27}}$

$s_p=4.237$$s_p=4.237$

Confidence interval$=\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}·s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

Confidence interval $=(16.23-17.69)\pm 2.052\times 4.237\sqrt{\frac{1}{15}+\frac{1}{14}}$

Confidence interval $=-1.46\pm 3.23$

Confidence interval $=-4.69to1.77$